Question

Let an ellipse and a hyperbola have same foci. If the length of conjugate axis of the hyperbola is equal to the length of minor axis of the ellipse, then the value of $\frac{1}{e_1^2}+\frac{1}{e_2^2}$ is $(e_1$ and $e_2$ denote the eccentricities of the two conics$)$

• A. $1$
• B. $4$
• C. $3$
• D. $2$

Answer 1

The correct option is D $2$

$\frac{x^2}{a_1^2}+\frac{y^2}{b_1^2}=1,a_1>b_1$

$\frac{x^2}{a_2^2}-\frac{y^2}{b_2^2}=1$

$(\pm a_1{e}_1,0)\equiv (\pm a_2{e}_2,0)$
$\Rightarrow a_1{e}_1=a_2{e}_2\cdots \left(1\right)$

Also, $2b_1=2b_2$
$\Rightarrow b_1=b_2$
$b_1^2=b_2^2$
$\Rightarrow a_1^2(1-e_1^2)=a_2^2(e_2^2-1)$
$\frac{a_2^2{e}_2^2}{e_1^2}(1-e_1^2)=a_2^2(e_2^2-1)\left[\text{From (1)}\right]$
$\Rightarrow \frac{1}{e_1^2}-1=1-\frac{1}{e_2^2}$
$\Rightarrow \frac{1}{e_1^2}+\frac{1}{e_2^2}=2$